Question

\((1-i\sqrt{3})^{2025}= \)

• A. \(2^{2025}\)
• B. \(2^{2026}\)
• C. \(-2^{2025}\)
• D. \(-2^{2026}\)

Hint:

When dealing with high powers of complex numbers, converting to polar form and using De Moivre's theorem is almost always the most efficient method. Remember that \( \cos(n\pi) \) is 1 for even n and -1 for odd n, while \( \sin(n\pi) \) is always 0 for integer n.

Answer 1

The Correct Option is C

First, we convert the complex number \(z = 1 - i\sqrt{3}\) into its polar form, \(r(\cos\theta + i\sin\theta)\).
The modulus is \(r = |z| = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2\).
The argument \(\theta\) is found from \(\cos\theta = 1/2\) and \(\sin\theta = -\sqrt{3}/2\). This places the angle in the fourth quadrant.
So, \(\theta = -\frac{\pi}{3}\).
The polar form is \(z = 2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3}))\).
Now, we use De Moivre's Theorem, which states \([r(\cos\theta + i\sin\theta)]^n = r^n(\cos(n\theta) + i\sin(n\theta))\).
\( (1-i\sqrt{3})^{2025} = 2^{2025} \left(\cos\left(2025 \cdot -\frac{\pi}{3}\right) + i\sin\left(2025 \cdot -\frac{\pi}{3}\right)\right) \).
Let's simplify the angle: \(2025 \cdot -\frac{\pi}{3} = -675\pi\).
Using the properties of cosine and sine: \(\cos(-x) = \cos(x)\) and \(\sin(-x) = -\sin(x)\).
The expression becomes \(2^{2025}(\cos(675\pi) - i\sin(675\pi))\).
Since 675 is an odd integer, \(675\pi\) is coterminal with \(\pi\).
\(\cos(675\pi) = \cos(\pi) = -1\).
\(\sin(675\pi) = \sin(\pi) = 0\).
Substituting these values back: \(2^{2025}(-1 - i \cdot 0) = -2^{2025}\).